AcWing 796. 子矩阵的和
题目描述
输入一个 n 行 m 列的整数矩阵，再输入 q 个询问，每个询问包含四个整数 x1,y1,x2,y2，表示一个子矩阵的左上角坐标和右下角坐标。
对于每个询问输出子矩阵中所有数的和。
输入格式
第一行包含三个整数 n，m，q。
接下来 n 行，每行包含 m 个整数，表示整数矩阵。
接下来 q 行，每行包含四个整数 x1,y1,x2,y2，表示一组询问。
输出格式
共 q 行，每行输出一个询问的结果。
数据范围
1≤n,m≤1000,
1≤q≤200000,
1≤x1≤x2≤n1,
1≤y1≤y2≤m1,
−1000≤矩阵内元素的值≤1000
输入样例：
cpp
3 4 3
1 7 2 4
3 6 2 8
2 1 2 3
1 1 2 2
2 1 3 4
1 3 3 4输出样例：
cpp
17
27
21思路
C++
cpp
#include <iostream>
using namespace std;
int main() {
int n, m, q;
scanf("%d%d%d", &n, &m, &q);
int s[n + 1][m + 1];
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
int tmp;
scanf("%d", &tmp);
s[i][j] = s[i][j - 1] + s[i - 1][j] - s[i - 1][j - 1] + tmp;
}
}
while (q--) {
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
printf("%d\n", s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1]);
}
}
cpp
#include <iostream>
using namespace std;
const int N = 1010;
int n, m, q;
int s[N][N];
int main() {
scanf("%d%d%d", &n, &m, &q);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
scanf("%d", &s[i][j]);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
s[i][j] += s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1];
while (q--) {
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
printf("%d\n", s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1]);
}
return 0;
}Go
go
package main
import (
"bufio"
"fmt"
"os"
)
func main() {
reader := bufio.NewReader(os.Stdin)
var n, m, q int
fmt.Fscan(reader, &n, &m, &q)
s := make([][]int, n+1)
for i := range s {
s[i] = make([]int, m+1)
}
for i := 1; i <= n; i++ {
for j := 1; j <= m; j++ {
var tmp int
fmt.Fscan(reader, &tmp)
s[i][j] = s[i-1][j] + s[i][j-1] - s[i-1][j-1] + tmp
}
}
writer := bufio.NewWriter(os.Stdout)
defer writer.Flush()
for i := 0; i < q; i++ {
var x1, y1, x2, y2 int
fmt.Fscan(reader, &x1, &y1, &x2, &y2)
result := s[x2][y2] - s[x1-1][y2] - s[x2][y1-1] + s[x1-1][y1-1]
fmt.Fprintln(writer, result)
}
}模板
cpp
S[i, j] = 第i行j列格子左上部分所有元素的和
以(x1, y1)为左上角，(x2, y2)为右下角的子矩阵的和为：
S[x2, y2] - S[x1 - 1, y2] - S[x2, y1 - 1] + S[x1 - 1, y1 - 1]